Question

sin(3A)/(sin(2A)-sin(A) = cos (??)

Answer 1

express the first formula with only "cos"

Answer 2

sin(3A) = sin(2A+A) (this is just the top)
= sin(2A)cos(A)+sin(A)cos(2A) (trig identity of sin(a+b))
= sin(A+A)cos(A)+sin(A)cos(A+A) (sin(2A) = sin (A+A))
= 2sin(A)cos(A)cos(A)+sin(A)[cos(A)^2-sin(A)^2] (trig identity of cos(a+b)
= 2sin(A)cos(A)^2 + sin(A)[(cos(A)^2 - (1- cos(A)^2)] (pythagorean identity sin^2+cos^2=1)
= sin(A) [2cos(A)^2 + [cos(A)^2 - (1-cos(A)^2)] (factoring out a sin(A))
= sin(A) [4cos(A)^2 - 1] (adding like terms)
= sin(A) [2cos(A) + 1][2cos(A) -1] (Factoring apart the 4cos(A)^2 - 1)

So the top ends up being sin(A)[2cos(A) + 1][2cos(A) - 1]

Now the bottom:
Sin(2A) - Sin(A) = 2sin(A)cos(A) - sin(A)
= sin(A)[2cos(A) - 1]

notice we have: sin(A) and [2cos(A) - 1] on the top and bottom:

sin(A)[2cos(A) + 1][2cos(A) - 1] / sin(A)[2cos(A) - 1]

= 2cos(A) + 1

Answer 3

wow thanks a lot !!